Question: Solve for $x$ : $ 8|x + 1| + 3 = -1|x + 1| + 10 $
Answer: Add $ {1|x + 1|} $ to both sides: $ \begin{eqnarray} 8|x + 1| + 3 &=& -1|x + 1| + 10 \\ \\ { + 1|x + 1|} && { + 1|x + 1|} \\ \\ 9|x + 1| + 3 &=& 10 \end{eqnarray} $ Subtract ${3}$ from both sides: $ \begin{eqnarray} 9|x + 1| + 3 &=& 10 \\ \\ { - 3} &=& { - 3} \\ \\ 9|x + 1| &=& 7 \end{eqnarray} $ Divide both sides by ${9}$ $ \dfrac{9|x + 1|} {{9}} = \dfrac{7} {{9}} $ Simplify: $ |x + 1| = \dfrac{7}{9}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x + 1 = -\dfrac{7}{9} $ or $ x + 1 = \dfrac{7}{9} $ Solve for the solution where $x + 1$ is negative: $ x + 1 = -\dfrac{7}{9} $ Subtract ${1}$ from both sides: $ \begin{eqnarray} x + 1 &=& -\dfrac{7}{9} \\ \\ {- 1} && {- 1} \\ \\ x &=& -\dfrac{7}{9} - 1 \end{eqnarray} $ Change the ${ - 1}$ to an equivalent fraction with a denominator of $9$ $ x = - \dfrac{7}{9} {- \dfrac{9}{9}} $ $ x = -\dfrac{16}{9} $ Then calculate the solution where $x + 1$ is positive: $ x + 1 = \dfrac{7}{9} $ Subtract ${1}$ from both sides: $ \begin{eqnarray} x + 1 &=& \dfrac{7}{9} \\ \\ {- 1} && {- 1} \\ \\ x &=& \dfrac{7}{9} - 1 \end{eqnarray} $ Change the ${ - 1}$ to an equivalent fraction with a denominator of $9$ $ x = \dfrac{7}{9} {- \dfrac{9}{9}} $ $ x = -\dfrac{2}{9} $ Thus, the correct answer is $x = -\dfrac{16}{9} $ or $x = -\dfrac{2}{9} $.